If m is chosen in the quadratic equation (m^2+1)x^2-3x+(m^2+1)=0 such that the sum of its …

Question

If $m$ is chosen in the quadratic equation $(m^2+1)x^2-3x+(m^2+1)=0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is:

Accepted Answer

Sum of roots $=\dfrac3{m^2+1}$ is greatest when $m=0$, giving $x^2-3x+1=0$, so $\alpha+\beta=3,\ \alpha\beta=1$.

$|\alpha-\beta|=\sqrt{9-4}=\sqrt5$.

$$|\alpha^3-\beta^3|=|\alpha-\beta|(\alpha^2+\alpha\beta+\beta^2)=\sqrt5(9-1)=8\sqrt5.$$

Solutions · 1